Pass char pointer/array to function

I'm trying to understand more about char pointers in C, but one thing got me.

Suppose I want to pass a char pointer to a function and change the value represented by that pointer. Here is an example:

int Foo (char *(&Msg1), char* Msg2, char* Msg3){
    char *MsgT = (char*)malloc(sizeof(char)*60);
    strcpy(MsgT,"Foo - TEST");
    Msg1 = MsgT; // Copy address to pointer
    strcpy(Msg2,MsgT); // Copy string to char array
    strcpy(Msg3,MsgT); // Copy string to char pointer
    return 0;
}

int main() {
    char* Msg1; // Initial char pointer
    char Msg2[10]; // Initial char array
    char* Msg3 = (char*)malloc(sizeof(char) * 10); // Preallocate pointer memory
    Foo(Msg1, Msg2, Msg3);
    printf("Msg1: %s\n",Msg1); // Method 1
    printf("Msg2: %s\n",Msg2); // Method 2
    printf("Msg3: %s\n",Msg3); // Method 3
    free(Msg1);    
    free(Msg3);
    return 0;
}

In the example above, I've listed all the working methods I know of to pass a char pointer to a function. What I don't understand is method 1 .

char *(&Msg1)What is the meaning of the first parameter passed to the function ?Foo

Also, method 2 and method 3 seem to be covered extensively by books and tutorials, some of which even refer to those methods as the most correct way to pass arrays/pointers . I don't know that method 1 looks fine to me, especially when I write an API where users can easily pass null pointers to functions without pre-allocating memory. If the user forgets to free the memory block ( same as method 3 ) , the only downside could be a potential memory leak . Is there any reason why we would prefer to use method 2 or 3 over method 3 ?